At 8: 00 am a bicyclist departed city a, heading towards city b at a rate of 20 km/h. after traveling for 4 hours he stopped for an hour break. at the same time, a biker left city a traveling in the same direction at the rate of 50 km/h. how long will it take the biker to catch up to the cyclist?

2 hours and 40 minutes

Step-by-step explanation:

After 4 hours, 1st person went 20 * 4 = 80 km

+1 hours rest

So now after total 5 hours (when 2nd one started), 1st one is 80 km ahead.

2nd bicyclist rate is 50

1st bicyclist rate is 20

So, for each hour, 2nd bicyclist advances 50 - 20 = 30 kilometers

How long will it take him, at this rate, to catch up 80 kilometers??

Simple!

30 km catch up every hour, so

80/30 = 2 2/3rd hours

That would be 2 hours 40 minutes to catch up

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!